模λ函数

模λ函数在由下式生成的模群（英语：Modular group）的主同余子群Γ(2)的作用下保持不变：^[3]:115

${\displaystyle \tau \mapsto \tau +2\ ;\ \tau \mapsto {\frac {\tau }{1-2\tau }}\ .}$ 

模群自身的生成元则以如下方式作用于模λ函数之上：^[3]:109

${\displaystyle \tau \mapsto \tau +1\ :\ \lambda \mapsto {\frac {\lambda }{\lambda -1}}\,;}$ 

${\displaystyle \tau \mapsto -{\frac {1}{\tau }}\ :\ \lambda \mapsto 1-\lambda \ .}$ 

λ函数为亚可比模量（Jacobi modulus）的平方^[3]:108，即${\displaystyle \lambda (\tau )=k^{2}(\tau )}$ ；亦可以戴德金η函数与Θ函数表达：

${\displaystyle \lambda (\tau )={\Bigg (}{\frac {{\sqrt {2}}\,\eta ({\tfrac {\tau }{2}})\eta ^{2}(2\tau )}{\eta ^{3}(\tau )}}{\Bigg )}^{8}={\frac {16}{\left({\frac {\eta (\tau /2)}{\eta (2\tau )}}\right)^{8}+16}}={\frac {\theta _{2}^{4}(0,\tau )}{\theta _{3}^{4}(0,\tau )}}}$ 

${\displaystyle {\frac {1}{{\big (}\lambda (\tau ){\big )}^{1/4}}}-{\big (}\lambda (\tau ){\big )}^{1/4}={\frac {1}{2}}\left({\frac {\eta ({\tfrac {\tau }{4}})}{\eta (\tau )}}\right)^{4}=2\,{\frac {\theta _{4}^{2}(0,{\tfrac {\tau }{2}})}{\theta _{2}^{2}(0,{\tfrac {\tau }{2}})}}}$ 

其中：^[3]:63

${\displaystyle \theta _{2}(0,\tau )=\sum _{n=-\infty }^{\infty }q^{\left({n+{\frac {1}{2}}}\right)^{2}}}$ 

${\displaystyle \theta _{3}(0,\tau )=\sum _{n=-\infty }^{\infty }q^{n^{2}}}$ 

${\displaystyle \theta _{4}(0,\tau )=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{n^{2}}}$ 

${\displaystyle q=e^{\pi i\tau }}$ 

λ函数亦可以魏尔斯特拉斯椭圆函数在定义其的格子的棱边中点和面心处的函数值表达；若令${\displaystyle [\omega _{1},\omega _{2}]}$ 为满足${\displaystyle \tau ={\frac {\omega _{2}}{\omega _{1}}}}$ 的基本周期二元组：

${\displaystyle e_{1}=\wp \left({\frac {\omega _{1}}{2}}\right),e_{2}=\wp \left({\frac {\omega _{2}}{2}}\right),e_{3}=\wp \left({\frac {\omega _{1}+\omega _{2}}{2}}\right)}$ 

则有：^[3]:108

${\displaystyle \lambda ={\frac {e_{3}-e_{2}}{e_{1}-e_{2}}}\,.}$ 

魏尔斯特拉斯函数在上述三点的值各不相同，这意味着λ函数取不到值0或1。^[3]:108

其与克莱因j函数（英语：Klein J-invariant）的关系为：^[3]:117[4]

${\displaystyle j(\tau )={\frac {256(1-\lambda (1-\lambda ))^{3}}{(\lambda (1-\lambda ))^{2}}}={\frac {256(1-\lambda +\lambda ^{2})^{3}}{\lambda ^{2}(1-\lambda )^{2}}}\ .}$ 

有一个与模λ函数相关的函数：λ*(x)函数，其给出了椭圆模量k的值。第一类完全椭圆积分K(k)与其互补对应的${\displaystyle K\left({\sqrt {1-k^{2}}}\right)}$ 关系如下：

${\displaystyle {\frac {K\left[{\sqrt {1-\lambda ^{*}(x)^{2}}}\right]}{K[\lambda ^{*}(x)]}}={\sqrt {x}}}$ 

λ*(x)函数的函数值可透过下列式子计算：

${\displaystyle \lambda ^{*}(x)={\frac {\vartheta _{2}^{2}[0;\exp(-\pi {\sqrt {x}})]}{\vartheta _{3}^{2}[0;\exp(-\pi {\sqrt {x}})]}}}$ 

${\displaystyle \lambda ^{*}(x)=\left[\sum _{a=-\infty }^{\infty }\exp[-(a+1/2)^{2}\pi {\sqrt {x}}]\right]^{2}\left[\sum _{a=-\infty }^{\infty }\exp(-a^{2}\pi {\sqrt {x}})\right]^{-2}}$ 

${\displaystyle \lambda ^{*}(x)=\left[\sum _{a=-\infty }^{\infty }\operatorname {sech} [(a+1/2)\pi {\sqrt {x}}]\right]\left[\sum _{a=-\infty }^{\infty }\operatorname {sech} (a\pi {\sqrt {x}})\right]^{-1}}$ 

其中${\displaystyle \vartheta }$ 为Θ函数。

此外λ函数与λ*(x)函数存在下列关联：

${\displaystyle \lambda ^{*}(x)={\sqrt {\lambda (i{\sqrt {x}})}}}$ 

所有的有理数r，${\displaystyle K\left(\lambda ^{*}(r)\right)}$ 与${\displaystyle E\left(\lambda ^{*}(r)\right)}$ 都可以视为椭圆积分的奇异值，可透过有限的伽马函数表示^[5]。

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